Computability and Logic

248 NORMAL FORMS
For (1) and (3.1) imply
∀x 1 ∀x 2 ∃y 2 R(x 1 , f 1 (x 1 ), x 2 , y 2 )
which with (3.2) implies (2). The sentences (3) are called the Skolem axioms.
For a prenex formula of a different kind, with different numbers of universal and
existential quantifiers, the number and number of places of the required Skolem functions
would be different, and the Skolem axioms correspondingly so. For instance, for
(1 ′ )
∃y 0 ∀x 1 ∀x 2 ∃y 1 ∃y 2 ∀x 3 Q(y 0 , x 1 , x 2 , y 1 , y 2 , x 3 )
we would need one zero-place function symbol (which is to say, one constant) f 0 and
two two-place function symbols f 1 and f 2 . The Skolem normal form would be
(2 ′ )
and the Skolem axioms would be
(3.0 ′ )
(3.1 ′ )
(3.2 ′ )
∀x 1 ∀x 2 ∀x 3 Q( f 0 , x 1 , x 2 , f 1 (x 1 , x 2 ), f 2 (x 1 , x 2 ), x 3 )
∃y 0 ∀x 1 ∀x 2 ∃y 1 ∃y 2 ∀x 3 Q(y 0 , x 1 , x 2 , y 1 , y 2 , x 3 ) →
∀x 1 ∀x 2 ∃y 1 ∃y 2 ∀x 3 Q( f 0 , x 1 , x 2 , y 1 , y 2 , x 3 )
∀x 1 ∀x 2 (∃y 1 ∃y 2 ∀x 3 Q( f 0 , x 1 , x 2 , y 1 , y 2 , x 3 ) →
∃y 2 ∀x 3 Q( f 0 , x 1 , x 2 , f 1 (x 1 , x 2 ), y 2 , x 3 ))
∀x 1 ∀x 2 (∃y 2 ∀x 3 Q( f 0 , x 1 , x 2 , f 1 (x 1 , x 2 ), y 2 , x 3 ) →
∀x 3 Q( f 0 , x 1 , x 2 , f 1 (x 1 , x 2 ), f 2 (x 1 , x 2 ), x 3 )).
But in exactly the same way in any example, the Skolem normal form will imply
the original formula, and the original formula together with the Skolem axioms will
imply the Skolem normal form.
If L is a language and L + is the result of adding Skolem functions for some or all of
its sentences, then an expansion M + of an interpretation M of L to an interpretation
of L + is called a Skolem expansion if it is a model of the Skolem axioms.
19.6 Lemma (Skolemization). Every interpretation of L has a Skolem expansion.
Proof: The essential idea of the proof is sufficiently illustrated by the case of our
original example (1) above. The proof uses a set-theoretic principle known as the
axiom of choice. According to this principle, given any family of nonempty sets,
there is a function ε whose domain is that family of sets, and whose value ε(X) for
any set Y in the family is some element of Y . Thus ε ‘chooses’ an element out of each
Y in the family. We apply this assumption to the family of nonempty subsets of |M|
and use ε to define a Skolem expansion N = M + of M.
We first want to assign a denotation f1 N that will make the Skolem axiom (3.1)
come out true. To this end, for any element a 1 in |M| consider the set B 1 of all b 1 in
|M| such that a 1 and b 1 satisfy ∀x 2 ∃y 2 R(x 1 , y 1 , x 2 , y 2 )inM.IfB 1 is empty, then no
matter what we take f1
N (a 1)tobe,a 1 will satisfy the conditional
∃y 1 ∀x 2 ∃y 2 R(x 1 , y 1 , x 2 , y 2 ) →∀x 2 ∃y 2 R(x 1 , f 1 (x 1 ), x 2 , y 2 )
since it will not satisfy the antecedent. But for definiteness, let us say that if B 1
is empty, then we are to take f1
N (a 1)tobeε(|M|). If B 1 is nonempty, then we